1. ## should be simple

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

I think the unit conversion is what keeps making my answer wrong....

2. Originally Posted by winterwyrm
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

I think the unit conversion is what keeps making my answer wrong....

Let's find out if it is the units.

Set up a coordinate system where the origin is where the car starts to brake and a +x direction in the direction of the velocity of the car at the origin.

This is what we know:
$\displaystyle t =$?
$\displaystyle x_0 = 0~ft$ $\displaystyle x =$?
$\displaystyle v_0 = 50~mi/h$ $\displaystyle v = 0~ft/s$
$\displaystyle a = -42~ft/s^2$

So let's first convert that initial speed:
$\displaystyle \frac{50~mi}{1~h} \cdot \frac{5280~ft}{1~mi} \cdot \frac{1~h}{3600~s} = 73.33~ft/s$

We want x. So
$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

$\displaystyle 0^2 = v_0^2 + 2ax$

$\displaystyle x = -\frac{v_0^2}{2a}$

I get x = 64.0212 ft.

-Dan

3. yup it was the units, I was trying to convert the other way around, and just threw a ratio in there, apparently in the wrong place.... Thanks a ton!

One thing though, where did you get that equation?

4. Hello, winterwyrm!

Another approach . . .

A car is traveling at 50 mi/hr when the brakes are fully applied,
producing a constant deceleration of 42 ft/sē.
What is the distance covered before the car comes to a stop?

Acceleration is the derivative of velocity.

We have: .$\displaystyle a(t) \:=\:v'(t) \:=\:-42$

Integrate: .$\displaystyle v(t) \:=\:-42t +C_1$

When $\displaystyle t = 0,\;v(0) \:=\:50\text{ mi/hr} \:=\:\frac{220}{3}\text{ ft/sec}$
. . So we have: .$\displaystyle \frac{220}{3} \:=\:-42(0) + C_1\quad\Rightarrow\quad C_1 \:=\:\frac{220}{3}$
. . Hence: .$\displaystyle v(t) \:=\:-42t + \frac{220}{3}$ . [1]

Integrate: .$\displaystyle s(t) \:=\:-21t^2 + \frac{220}{3}t + C_2$

When $\displaystyle t = 0,\;s(0) \:=\:0$
. . So we have: .$\displaystyle 0 \:=\:-21(0^2) + \frac{220}{3}(0) + C_2\quad\Rightarrow\quad C_2 \:=\:0$
. . Hence: .$\displaystyle s(t) \;=\;-21t^2 + \frac{220}{3}t$ . [2]

The car comes to a stop when $\displaystyle v(t) = 0.$
. . From [1]: .$\displaystyle -42t + \frac{220}{3} \;=\;0\quad\Rightarrow\quad t \;=\;\frac{110}{63}$ seconds.

Substitute into [2]: .$\displaystyle s\!\left(\frac{110}{63}\right) \;=\;-21\left(\frac{110}{63}\right)^2 + \frac{220}{3}\left(\frac{110}{63}\right) \;=\;\frac{12,100}{189} \;\approx\; 64.0212\text{ ft}$